Question

Define electric potential due to a point charge and arrive at the expression for the electric potential at a point due to a point charge in its vicinity.

Answer 1

Hint: At infinite distance when an electric charge is not under the influence of the point charge, the potential is defined as zero. Electric potential at a given position is the amount of work that has to be done to bring the charge at a finite distance from the point charge.

Formula Used:
Coulomb's law gives the force of attraction between two charged particles. The mathematical expression is,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Qq}{{{r}^{2}}}$...........................(1)

Where,
${{\varepsilon }_{0}}$ is the di-electric constant of space
$Q$ is the charge of one particle
$q$ is the charge of the other particle
$r$ is the distance between two particles.

Complete step by step answer:
The electrostatic potential at a point is measured by the work done to bring a unit charge from infinite distance to that point in the electric field.

Let’s define the system in the following diagram.

Let’s assume there is a positive charge +Q and we need to find the electric potential at P which is at a distance of r from the charge.

As we need to find the work done of a variable force, we need to take the help of integration.

Work done is given by,
$W=\int{F.dx}$

Where,
$F$ is the force at that location
$dx$ is the infinitesimal distance

Let’s assume that the unit charge is at a distance ‘x’ from the point charge.

Hence, we can find the force of repulsion using Coulomb’s law,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{x}^{2}}}$

$Q$ is the charge of the point charge
$x$ is the distance from the point charge
${{\varepsilon }_{0}}$ is the di-electric constant of space
Let’s assume $dx$ is an infinitesimal distance towards the point charge.

As we are working against the force, work done will be positive.

Hence, we can write,
$W=\int\limits_{\infty }^{r}{(\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{x}^{2}}})dx}$
$\Rightarrow W=-[\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{x}]_{\infty }^{r}$
$\Rightarrow W=-[-\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{r}]$
$\Rightarrow W=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{r}$

So, the electrostatic potential at a distance r from a point charge (+Q) is given by,
$U=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{r}$

Note: Both the point charge and the unit charge are positive. So, we had to work against the electrostatic force. Hence, the potential is positive. If, however, the point charge had negative value, we had to work along the force. So, in that case potential would have been negative.